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The karnaugh map method, a diagrammatic technique used to simplify boolean functions. It covers two-variable, three-variable, four-variable, and five-variable maps, as well as the product of sums simplification and don't-care conditions. The document also discusses nand and nor gate implementations.
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Karnaugh Maps A Karnaugh map is a diagram made up of squares used for the purpose of simplifying Boolean functions. The Karnaugh map is like a graphical representation of a truth table. Each square in the map represents one minterm of a Boolean function. The map represents all possible ways a function can be expressed in standard form. Any two adjacent squares differ by only one variable which is primed in one square and unprimed in the other.
Two and Three-Variable Maps A generic two-variable map Representation of functions in a two variable map. example: f 1 = xy example: f 2 = x + y x\y y’ y x’ x’y’ x’y x xy’ xy x\y y’ y x’ m 0 m 1 x m 2 m 3 x\y y’ y x’ 0 0 x 0 1 x\y y’ y x’ 0 1 x 1 1 x y f 1 minterm 0 0 0 m 0 0 1 0 m 1 1 0 0 m 2 1 1 1 m 3 x y f 2 minterm 0 0 0 m 0 0 1 1 m 1 1 0 1 m 2 1 1 1 m 3 x y minterm 0 0 m 0 0 1 m 1 1 0 m 2 1 1 m 3
example: Simplify the Boolean function F ( x , y , z ) = x’z’ + xyz + xz’. F ( x , y , z )= (0, 2, 4, 6, 7) Step 1 – grouping Make a group of four by grouping m 0 , m 2 , m 4 , and m 6 = z’. Make a group of two by grouping m 7 and m 6 = xy. Step 2 – read the terms from each group and sum them F ( x , y , z )= xy + z’ Why does a Karnaugh map work? A Karnaugh map works because two adjacent squares differ by only one variable. When these two squares are grouped, the variable that differs will be eliminated because any variable ORed with its complement is 1. The number of variables eliminated from a term is related to the size of the group. Three-variable Map size of square literals eliminated 1 0 2 1 4 2 8 3 example from a three-variable map: m 1 + m 5 = x’y’z + xy’z = y’z ( x’ + x ) = y’z x\yz y’z’ y’z yz yz’ x’ 1 0 0 1 x 1 0 1 1
example: Simplify the Boolean function F ( A , B , C ) = A’BC’ + AB + A’BC. F ( A , B , C ) = (2, 3, 6, 7) Step 1 – grouping Make a group of four by grouping m 2 , m 3 , m 6 , and m 7 = B. Step 2 – read the terms from each group and sum them F ( x , y , z )= B example: Simplify the Boolean function F ( A , B , C ) = (0, 5, 7). Step 1 – grouping Make a group of one with m 0 = A’B’C’. Make a group of two by grouping m 5 and m 7 = AC. Step 2 – read the terms from each group and sum them F ( x , y , z )= A’B’C’ + AC example: Simplify the Boolean function F ( A , B , C ) = AB’ + AB’C. AB’ = AB’ ( C + C’ ) = AB’C + AB’C’ F ( A , B , C ) = (4, 5) Step 1 – grouping Make a group of two by grouping m 4 and m 5 = AB’. Step 2 – read the terms from each group and sum them F ( A , B , C )= AB’
example: Simplify the Boolean function F ( A , B , C , D ) = A’B’ CD + ABCD’ + A’BD’
example: Find the prime implicants and essential prime implicants for the Boolean function F ( A , B , C , D ) = (1, 3, 5, 7, 8, 9, 10, 11, 12, 13). Step 1 – find largest possible groups group 1: m 1 , m 3 , m 5 , and m 7 = A’D group 2: m 1 , m 3 , m 9 , and m 11 = B’D group 3: m 12 , m 13 , m 8 , and m 9 = AC’ group 4: m 1 , m 5 , m 13 , and m 9 = C’D group 5: m 8 , m 9 , m 11 , and m 10 = AB’ Step 2 – identify essential prime implicants essential prime implicants: A’D , AC’ , AB’ Step 3 – identify prime implicants prime implicants: B’D , C’D
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example: Simplify the Boolean function F ( A , B , C , D , E ) = (0, 2, 4, 6, 9, 13, 21, 23, 25, 29, 31). Step 1 – grouping Make a group of four by grouping m 0 , m 4 , m 2 , and m 6 = A’B’E’. Make a group of four by grouping m 21 , m 23 , m 29 , and m 31 = ACE. Make a group of four by grouping m 13 , m 9 , m 29 , and m 25 = BD’E. Step 2 – read the terms from each group and sum them F ( A , B , C , D , E ) = A’B’E’ + ACE + BD’E
example: Simplify the Boolean function F ( A , B , C , D , E ) = (1, 3, 9, 11, 13, 29). Step 1 – grouping Make a group of four by grouping m 1 , m 3 , m 9 , and m 11 = A’C’E. Make a group of two by grouping m 13 , and m 29 = BCD’E. Step 2 – read the terms from each group and sum them F ( A , B , C , D , E ) = A’C’E + BCD’E
The minimized Boolean functions derived from the maps in all previous examples were expressed in sum of products form. With a minor modification, the product of sums form can be obtained. The 1s placed in the squares of the map represent the minterms of the function F. The 0s represent the minterms of the complement of the function, i.e., F’. The complement of F’ is the original function F. Because of the generalized DeMorgan’s theorem, the function so obtained is automatically in the product of sums form.
For some functions, not all input combinations are valid, e.g., a function who’s input will be coded in BCD. (BCD uses only 10 of the possible 16 combinations) The minterms associated with these invalid or unused inputs are called “don’t- care conditions” Including don’t-cares in a Karnaugh map is a way to achieve further simplification Don’t-cares are represented on a map with an “X” Individual don’t-cares are read as either a zero or a one, whichever results in the greater simplification Example: F ( A,B,C,D) = (0, 2, 4, 6, 8) This is a function to light an LED if an even number is input. The input will be in BCD so minterms 10 , 11 , 12 , 13 , 14 , and 15 are don’t-cares. First, without the don’t-cares AB\CD C’D’ C’D CD CD’ A’B’ 1 0 0 1 A’B 1 0 0 1 AB 0 0 0 0 AB’ 1 0 0 0 F ( A,B,C,D) = A’D’ + B’C’D’ Without don’t-cares
Now, with the don’t-cares AB\CD C’D’ C’D CD CD’ A’B’ 1 0 0 1 A’B 1 0 0 1 AB X X X X AB’ 1 0 X X F ( A,B,C,D) = D’ With don’t-cares Now, the same problem but we will simplify F’ AB\CD C’D’ C’D CD CD’ A’B’ 1 0 0 1 A’B 1 0 0 1 AB X X X X AB’ 1 0 X X F’ ( A,B,C,D) = D F ( A,B,C,D) = D’
NAND Implementation Requires the Boolean function to be simplified in SOP form Method 1 (two logic levels)
Example: F = (0, 2, 4, 6, 7) Method 1, NAND logic A\BC B’C’ B’C BC BC’ A’ 1 0 0 1 A 1 0 1 1 F = AB + C’ AND-OR-INVERT Logic NAND Logic
